Answer:
O B. Experimental group
Explanation:
An experimental group is a test sample or the group that receives an experimental procedure. This group is exposed to changes in the independent variable being tested.
Answer:
Percentage by mass of calcium carbonate in the sample is 93.58%.
Explanation:
Assumptions:
- calcium carbonate was dissolved completely and the amount of carbon dioxide released was proportional to the amount of calcium carbonate.
- the sample did not contain any other compound that released or reacted with carbon dioxide.
given the equation of the reaction:
from assumption:
amount carbon dioxide = amount of calcium carbonate
reacting mass ( m ) = Molar Mass ( M ) * Amount ( n )
m = 0.04938397 
* 100 
= 4.93839 
percentage by mass of CaC
= 
.
Answer:
Phosphatidylcholine also known as lecithin
Explanation:
The most common phospholipids in order of abundance are
1). phosphatidylcholine, lecithin
2). phosphatidylethanolamine, cephalins 3).phosphatidylinositol, and 4).phosphatidylserine.
They have the common characteristics of esterified fatty acids to the 1 and 2 positions of the glycerol structure with the esterified phosphate group to the 3 position
Answer:
Increase is the answer
Explanation:
Increase is the answer hopes this helps you
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
